# How do you find the intercept and vertex of y= (2x-1)(3x+4)?

Mar 6, 2016

Analyse the equation to find $y$ intercept $\left(0 , - 4\right)$, $x$ intercepts $\left(- \frac{4}{3} , 0\right)$, $\left(\frac{1}{2} , 0\right)$ and vertex $\left(- \frac{5}{12} , - \frac{121}{24}\right)$

#### Explanation:

If $y = 0$ then $\left(2 x - 1\right) = 0$ or $\left(3 x + 4\right) = 0$

Hence $x = \frac{1}{2}$ or $x = - \frac{4}{3}$

If $x = 0$ then $y = \left(0 - 1\right) \left(0 + 4\right) = - 4$

So the $x$ intercepts are $\left(- \frac{4}{3} , 0\right)$ and $\left(\frac{1}{2} , 0\right)$

and the $y$ intercept is $\left(0 , - 4\right)$

Since parabolas are bilaterally symmetric, the vertex will have $x$ coordinate exactly midway between the two $x$ intercepts:

$x = \frac{- \frac{4}{3} + \frac{1}{2}}{2} = \frac{- \frac{8}{6} + \frac{3}{6}}{2} = - \frac{5}{12}$

Substitute this value of $x$ back into the original equation to find:

$y = \left(2 \left(- \frac{5}{12}\right) - 1\right) \left(3 \left(- \frac{5}{12}\right) + 4\right)$

$= \left(- \frac{5}{6} - \frac{6}{6}\right) \left(- \frac{15}{12} + \frac{48}{12}\right)$

$= \left(- \frac{11}{6}\right) \left(\frac{33}{12}\right)$

$= \left(- \frac{11}{6}\right) \left(\frac{11}{4}\right)$

$= - \frac{121}{24}$

So the vertex is at $\left(- \frac{5}{12} , - \frac{121}{24}\right)$

graph{(2x-1)(3x+4) [-6.58, 5.904, -5.27, 0.97]}